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Centres at: MUKHERJEE NAGAR MUNIRKA UTTAM NAGAR DILSHAD GARDEN ROHINI BADARPUR BORDER
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SSC MAINS TEST - 13 (T-II) 2013 (SOLUTION)SSC MAINS TEST - 13 (T-II) 2013 (SOLUTION)SSC MAINS TEST - 13 (T-II) 2013 (SOLUTION)SSC MAINS TEST - 13 (T-II) 2013 (SOLUTION)
1. (*)7
5,
11
8,
3
2,
13
9
=4013
409
, 1803
1802
, 4511
458
, 727
725
=520
360,
540
360,
495
360,
504
360
In the above fractions, numerators are
equal, so the fraction with the greater
denominator will have the least value of
ratio and the fraction with the lowest value
of denominator will have the highest value
of ratio.
540
360<
520
360 <
504
360<
495
360
3
2<
13
9<
7
5 <
11
8
[The required ascending order of the
given fractions] Ans.
2. (D) 21
2
1
)347()31028(
1 1
2 2 2 22 25 ( 3) 2.5. 3 2 ( 3) 2.2. 3
5 3 (2 + 3 )1
5 3 32
1
5 3 34
32
5 3 (2 3 )
5 3 2 + 3 = 3 Ans
3. (B) Let x be a +ve integer.
Now, for x = 1
(A)x
x= 1,
(B)x
x 1=
1
2 = 2
(C)1x
x=
2
1= 0.5
and (D)3
2
x
x=
4
3 = 0.75
(B)x
x 1has the greatest value. Ans.
4. (D)Let the total number of swans be x.
The number of swans playing on the shore
of the pond = x2
7
Number of swans inside the pond = 2
x = 22
7x
2(x 2) = x7
4(x2 4x + 4) = 49x
4x2 16x + 16 49x = 0
4x2 65x + 16 = 0
On solving x = 16
Number of swans = 16 Ans.
5. (B)342
784
342
)7( 283
342
)343(28
According to remainder theroem,
342
)343(28
will have the same remainder
as 342
)1( 28
1 Ans.
6. (A) Let x represents number of students & yrepresents the number of rows.
Then,
No. of students in each row = y
x
Case : (I)
4
y
x (y 2) = x
2y2 4y = x ... (i)
Case : (II)
4
y
x (y + 4) = x
y2 + 4y = x ... (ii)
From eqn (i) & (ii)
2y2 4y = y2 + 4y
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2
y(y 8) = 0
y = 8
Total no. of students
x= 2(8)2 4 8
= 128 32
x = 96 Ans.
7. (C)Let the numbers be x and 4x.
Then, 84 21 = x 4x
4x2 = 1764
x2 = 441
x = 21Hence, the largest number is 4x = 84. Ans.
8. (A) Total one-digit number = 9, No. of digits =9
Total two-digit number = 90, No. of digits = 180
Total three-digit number = 900, No. of digits = 2700
Let there are x-4 digits numbers.
According to question,
90 + 180 + 2700 + 4 x = 3189
x = 4
28893189 = 75
Number of pages
= 9 + 90 + 900 + 75 = 1074 Ans.
9. (*) Sn
= 2
n[2a + (n 1)d]
S40
= 20[24 + 39 4]
= 20 164 = 3280 Ans.
10. (A) nth term of a GP = arn 1
8th term = 5 (2)8 1
= 5 (2)7
= 5 128
= 640 Ans.
11. (A)
Total students who are studying at leastone subject = 21 + 1 + 2 + 1 + 9 + 3 + 8
= 45
Students who are not studying any of thethree subjects. = 80 45 = 35 Ans.
12. (C) Let the average of 12 innings be x.
Also, 13
9612 x = x + 5
12x + 96 = 13x + 65
x = 31
His new average
= 13
963112
= 13
468= 36 Ans.
13. (A) No. of workers = x, No. of officers = y.
Case : (I)
x + y = 400 ... (i)
Case : (II)
2000 x + 10000 y = 400 3000
x + 5y = 2000
3000400
x + 5y = 600 ... (ii)Substracting (ii) from (i)
(x + 5y) (x + y) = 200
y = 50
No. of officers = y = 50
No. of workers = x = 400 50 = 350 Ans.
14. (A) Let their ages be x and y.
x
1 +
y
1= 5
yx
11
y + x = 5(y x) 6x = 4y
y
x=
3
2... (i)
Now, yx
xy
= 1
4.14
xy = 14.4(x + y) ... (ii)From Eqs. (i) and (ii),
x = 24 yrs. and y = 36 yrs.
15. (D) By the rule of alligation,
Quantity of rice sold at 10% gain
= 12
12 3 50 = 40 kg
Quantity of rice sold at 5% loss
= 312
3
50 = 10 kg Ans.
16. (A) Quantity of nitric acid = 10 10
1
= 1 litre Water = 10 1 = 9 litre
Suppose x litre water is added in thesolution.
Then, (10 + x) 100
4= 1
x = 15 litre Ans.
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17. (A) Let the average speed be x km/h.
Time taken by aircraft (t) = x
800
ATQ,
t 60
40=
40
800
x
x
800
40
800
x =
3
2
)40(
32000
xx = 3
2
x(x + 40) = 48000 x = 200 km/h Ans.
18. (C) Mohan can reach the middle in 12.5 min.
Puran can reach the middle in 25 min.
So, required time = 25 12.5
= 12.5 min Ans.
19. (D) Speed of X = 2
3Y
Distance covered by Y before catching = D m
2
3D = D + 300
2
1D = 300
D = 600 mTotal distance = 600 + 300 = 900 m Ans.
20. (C) Selling price of first shopkeeper
= 700 100
70
100
94 = ` 460.60
Selling price of second shopkeeper
= 700 80
100
84
100 = ` 470.40
Required difference
= 470.40 460.60 = ` 9.80
21. (D) Let the original value of fridge be ` x.
Then, cost price = 16
15x
Selling price = 100
110x
Gain percent = 100
16
1516
15
100
110
x
xx
= 17.33% Ans.
22. (D) Let the rate of flow of river be x km/h.
Then, downward speed = (10 + x) km/h.
ATQ,
)10(
91
x + )10(
91
x = 20
On solving, we have x = 3.
Hence, speed of flow of river is 3 km/h.
23. (A) Let the rate of filling tank be x m3 /min.
Then, the rate of empting the tank
= (x + 10) m3/min
ATQ,
x
2400
10
2400
x= 8
2400
)10(
10
xx= 8
x(x + 10) = 3000
x = 50 m3/min Ans.
24. (C) Let first pipe can fill the tank in x h.
Second pipe can fill the tank in (x 5) h
Third pipe can fill the tank in (x 9) h
ATQ,
x
1 +
5
1
x =
9
1
x
5
)5(
xx
xx = x 9
x2 5x = 2x2 23x + 45
(x 15)(x 3) = 0
x = 15 h as x = 3 h is not possible. Ans.
25. (B) Number of men = 5
2 25 = 10
Number of women= 5
3 25 = 15
Amount distributed among men and women
= 275 80%
= ` 220
Let the wages paid to a man be ` 5x and
to a woman be ` 4x, then
10 5x + 15 4x = 220
50x + 60x = 220
x = 2
Wages recieved by a woman= 2 4 = ` 8 Ans.
26. (C) 10% of profit = 100
10 800 = ` 80
Remaining Rs. 720 is divided in the ratio
= 1500 : 900 = 5 : 3
Share of Y =8
3 720 + 80 = ` 350 Ans.
27. (A)According to question,
%5%
2
18 of sum = 350
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sum = 5.3
350 100
= ` 10000 Ans.
28. (D) Cash price, CP = ` 39000
Cash down payment, DP = ` 17000
Balance due, after Ist instalment,
BD = ` 22000
P = value of instalment = ` 4800
n = no. of instalments = 5
R = rate of interest
ATQ,
12001
nRBD =
2400
)1(1
RnnP
1200
51
R22000 =
2400
41
R5 4800
1200
51
R11=
2400
41
R12
11 + 1200
55R = 12 +
1200
24R
1200
55R
1200
24R = 1
1200
31R = 1
R = 31
1200 = 38.71% Ans.
29. (B)
Profit received by Anu = ` 110 + Rs. 350
= ` 460
& Profit received by Bimla = ` 420
30. (A)Required % = 5 + 5 + 100
55
= 10.25% Ans.
31. (B)
Percentage of failed students
= 30% + 35% 27% = 38%
Percentage of passed students
= 100% 38% = 62%
Now, Let total number of students be x.
62% of x = 248
x = 248 62
100 = 400 Ans.
32. (C) Let the maximum marks be x.
Then,
296 259 = 5% of x
100
5x = 37
x = 740 Ans.
33. (C) Let Ram's monthly income be ` x.
Total savings = x 100
80
100
85
100
70
x = 9520 80
100
85
100
70
100
= ` 20000
34. (B) Total candidates = 2000
No. of boys = 900
No. of girls = 1100
No. of students who passed
= 100
90032 +
100
110038
= 288 + 418 = 706
No. of students who failed
= 2000 706 = 1294
Required percentage
= 2000
1294 100 = 64.7% Ans.
35. (D) Only the option (d) gives the difference of
votes between two candidates as 308. Ans.
36. (B) Let the cost price of colour printer and
computer system be ` x and ` y respectively.
According to question,
x 100
120 + y
100
90= x + y
0.2x = 0.1y ..... (i)
x 100
85 + y
100
105= x + y 800
0.05y= 0.15x 800
From Eqs. (i) and (ii),
x = ` 16000 Ans.
37. (D) Let the cost price of book be ` x.
Then, (1.2x 18) 0.8x = 0.25 0.08x
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0.4x 18 = 0.20x
x = 20.0
18= ` 90 Ans.
38. (A)Cost price of transistor = ` 320
Selling price of transistor
= 320 1.15 = ` 368
Marked price of transistor
= 368 + 32 = ` 400
Required percentage of profit
= 320
320400 100 = 25% Ans.
39. (A) Let the cost price of 12 apples be ` 100.
Then, selling price of 100 apples will be ` 800.
CP of 1 apple = 12
100= `
3
25
and SP of 1 apple = 100
800= ` 8
Loss =
8
3
25 = `
3
1
Loss percent = CP
Loss 100
=
3
253
1
100
= 3
1
25
3 100 = 4% Ans
40. (A) Let the distance covered by the first train
be x km/h.
As both trains have travelled for same
time.
50
x=
60
120x
60x = 50x + 6000
x = 600
Total distance = x + (x + 120)
= 1320 km Ans.
41. (D) Let the speed of train be x km/h.
As both the persons are walking in the
same direction of train.
So, (x 4.5) 8.4 = (x 5.4) 8.5
0.1x = 8.1
x = 81 km/h Ans.
42. (A) Let Ashokan can finish the work in x days.
Then,
Nitin can finish the work in 3x days.
3x x = 40
x = 20 days
and 3x = 60 days
So, together they can finish the work in
6020
6020 days = 15 days Ans.
43. (D) Men : Women : Boys = 15 : 24 : 36
= 5 : 8 : 12
Convert women and boys in terms of men
8 women= 5 men
12 women = 8
512 =
2
15men
12 boys = 5 men
6 boys =12
5 6 =
2
5men
Total women and boys in terms of men
=12
5 +
2
5 =
2
20 = 10 men
Let the number of men required = x
Then, (x + 10) =630
25.281215
= 18
x + 10 = 18
x = 8 men Ans.
44. (B) Let they make x pieces per day.
Then,x
360
4
360
x= 1
360
)4(4
xx= 1
x(x + 4) = 1440 = 36 40 x = 36
Required number of days
= 36
360= 10 days Ans.
45. (C) LCM of (3 + 2), (7 + 3)
and (11 + 4) is 30.
Let the capacity of each container be 30 l
Quantity of milk after mixing
=
15
11
10
7
5
3 30 = 61 l
Quantity of water after mixing
=
15
4
10
3
5
2 30
= 29 l
Required ratio = 61 : 29 Ans.
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46. (D) As the sum of money that are to be divided
among A, B and C and between E and F
are not given. So, the amount that B
receive cannot be determined. Ans.
47. (D) New ratio = 3 100
150 : 5
100
160 : 7
100
150
= 2
9 : 8 :
2
21 = 9 : 16 : 21 Ans.
48. (B)18 carat gold = 4
3 pure gold =
4
3 24 = 18
20 carat gold = 6
5 pur gold =
6
5 24 = 20
Required ratio = 18 : 20 = 9 : 10 Ans.49. (D) Let the income of two persons be ` 4x and
` 5x and their expenses be ` 7y and ` 9y
respectively.
Then,
4x 7y = 50 .... (i)
and 5x 9y = 50 .... (ii)
On solving Eqs. (i) and (ii), we get
x = 100 and y = 50
The income of persons are ` 400 and` 500. Ans.
50. (C)
Hence, the next number of the series will
be 27. Ans.
51. (B)8
x=
23
32
On applying componendo and dividendo:-
8
8
x
x=
23
233
... (i)
Again,12
x=
23
22
12
12
x
x=
32
323
... (ii)
From Eqs. (i) + Eqn (ii)
8
8
x
x +
12
12
x
x=
23
233
23
323
= 23
323233
= 23
)23(2
= 2 Ans.
52. (A) a = 3 + 22
a
1=
223
1
+
223
223
= 89
223
= 3 22
aa
1 = 6
3
246 1
a
aaa = 3
3 11
aaaa
=
3
3 1
aa +
aa
1
=
31
aa
aa
13 +
aa
1
63 3 6 + 6 = 216 18 + 6 = 204 Ans.53. (A) x = 997, y = 998, z = 999
Now,
zxyzxyzyx 222 =
})()(){(2
1 222 xzzyyx
= 2
1{(1)2 + (1)2 + (2)2}
= 2
1(1 + 1 + 4) = 3 Ans.
54. (D)a
a3
15 = 5
Multiplying both sides by 5
3
aa
5
13 = 3
Squaring both sides
aa
aa
5
132
25
19
2
2 = 9
22
25
19
aa = 9
5
6 =
5
39Ans.
55. (C) 22 1
xx + 2 = 3
22 1
xx = 1
x4 + 1 = x2
x4 x2 + 1 = 0
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Now, x206 + x200 + x90 + x84 + x18 + x12 + x6 + 1
= x200(x6 + 1) + x84(x6 + 1) + x12(x6 + 1) + 1(x6 + 1)
= (x6 + 1)(x200 + x84 + x12 + 1)
= (x2 + 1) (x4 x2 + 1) (x200 + x84 + x12 + 1)
0 Ans.
56. (A) a3 + b + c3 3abc
=
=
=2
1(32 + 52 + 12)
=2
1 35 = 17.5 Ans.
57. (A) If 4
2
2
2 ytx
y
x is a perfect square
It must be equal to
22
2
22y
y
xy
y
x
= xy
y
x
22
2
On comparing, we have
tx = x
t = 1 Ans.58. (D) 8x3 ax2 + 54x + b
(2x 3)3 = 8x3 3 2x 3(2x 3) 27
= 8x3 36x2 + 54x 27
On comparing with the given expression
a = 36, b = 27 Ans.
59. (B) a = 3b 7 .... (i)
a 3b = 7
a + 3 = 3(b + k) 7
a + 3 = 3b 3k 7
a 3b = 3k 10
7 = 3k 10
k = 1 Ans.
60. (C) y = |x| 5
|x| y 5 = 0
Area of two traingles formed by
|x| y 5 = 0
=
ab2
c2
2
= ab
c 2=
11
)5( 2
= 25 Sq. unit Ans.
61. (D)
In APY and BPX
X = Y = 90A tangent is always perpendicular to the
radius through the point of contact.
APY ~ BPX [By AA similarity]
BX
AY=
PB
AP
2
5=
PB
AP
P divides AB externally in the ratio of 7 : 2.
62. (D) DE || BC (given)
ADE and ABC are similar.
)(
)(
ABCar
ADEar =
2
AB
AD
2
1=
AB
AD
AB
AD=
2
1
BD
AD=
12
1
Ans.
63. (C) Length of the common tangent
= 22 )36( d
8 = 22 )36( d
or d2 = 64 + 81 = 145
d = 145 cm Ans.
Distance between their centres = 145 cm
64. (B) Let the number of sides in polygon = n
The sum of interior angles
= (n 2) 180
ATQ,
5 172 + (n 5) 160= 180n 360
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860 + 160n 800 = 180n 360
60 + 360= 180n 160n
20n = 420
n = 21 Ans.
65. (C) The sum of any two sides of a triangle
is greater than the 3rd side.
2 + 3 > 5, which is wrong.
2 + 3 > 6, which is wrong.
(2, 3, 5) or (2, 3, 6) will not form a triangle.Triplets (3, 5, 6) & (2, 5, 6) are true for the
sides of a triangle.
= 2 triangles Ans.
66. (C)
In ADE,AE = AB AE = AD
E = D = (say)
A = + = 2
Similarly, In BCF,BF = AB BF = BC
C = F = (say)
B = 2
In OEF,
A + B = 1802 + 2 = 180
+ = 90
EOF = 90
ED CF Ans.67. (B) BC = AD
Distance= BC
In AOB,
tan 30= AO
350
3
1=
AO
350
AO = 150 m
AD = 300 m
Speed = 2
300= 50 m/min
= 2
5 m/s
= 9 km/h Ans.
68. (B)Let the shortest side be x m.
The hypotenuse = (2x + 3)m
Let third side = y m
x + y + 2x + 3 = 6x y = (3x 3) mNow, (x)2 + (3x 3)2 = (2x + 3)2
x2 + 9x2 + 9 18x = 4x2 + 9 + 12x 6x2 30x = 0 x = 5 mas x 0 Three sides are 5 m, 12 m and 13 m.
69. (B)In AQC,
AQ2 = AC2 + CQ2
In PCB, BP2 = PC2 + CB2
AQ2 + BP2 = AC2 + CQ2 + PC2 + CB2
= (AC2 + CB2) + (CQ2 + PC2)
= AB2 + PQ2
= AB2 +
21
2AB
AQ2 + BP2 = 4
ABAB4 22
4(AQ2 + BP2) = 5AB2 Ans.70. (B) OABC is a rhombus with centre O.
Let diagonal of the rhombus be
OB = 2x and AC = 2y
Radius of the circle = OB = OA = OC = 2x
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In POC,
OC2 = OP2 + PC2
(2x)2 = x2 + y2
4x2 = x2 + y2
3x2 = y2 x = 3
y
Also, area of rhombus
2
1 (2x)(2y) = 332
x = 4
Radius of circle = 2 4 = 8 m Ans.
71. (C) a sin + b cos = cOn squaring,
a2 sin2 + b2 cos2 + 2ab sin cos = c2
...(i)
Let a cos b sin = k
a2 cos2 + b2 sin2 2ab sin cos= k2
... (ii)
From equation (i) & (ii),
a2(sin2 + cos2 ) + b2(sin2+ cos2 ) = c2 + k2
k2 = a2 + b2 c2
k = 222 cba Ans.
72. (D) sec =x
x
4
14 2
b = 4x, h = 4x2 + 1
Then,
P = 222 )4()14( xx
= 224 161816 xxx
= 1816 24 xx
= 4x2 1
sec + tan = x
x
4
14 2 +
x
x
4
14 2
= x
xx
4
1414 22
= x
x
4
8 2= 2x Ans.
73. (B) cos
sin1
1
sin1
1= 4
cos
2sin1
sin1sin1 = 4
cos
2cos
2= 4
cos =2
1
cos = cos 60
= 60 Ans.
74. (D) un
= cosn + sinn
Now,
2u6 3u
4 + 1
= 2(cos6 + sin6 ) 3(sin4 + cos4 ) + 1
= 2(cos2 + cos2 )3 3sin2 cos2
(sin2 + cos2 ) 3{(sin2 + cos2 )2
2sin2 cos2 )} + 1
= 2 6 sin2 .cos2 + 6sin2 cos2 3 + 1
= 2 3 + 1 = 0 Ans.
75. (B) (a2 b2) sin + 2ab cos = a2 + b2
22
22
ba
ba
sin + 22
2
ba
ab
cos = 1
Let 22
22
ba
ba
= sin & 22
2
ba
ab
= cos
Then, above equation becomes
sin2 + cos2 = 1
sin = 22
22
ba
ba
cos = 222
ba
ab
tan = ab
ba
2
22 =
1
2ab(a2 b2) Ans.
76. (D)
2
4
cos
cos +
2
4
sin
sin = 1
By taking
= , it satisfies the above equation
2
4
cos
cos +
2
4
sin
sin = 1 Ans.
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77. (B) l cos2 + m sin2 =
2
2
cot
cos
2
2
sin
sin1
l cos2 + m m cos2 = 1 + 1 cos2
l cos2 + cos2 m cos2 = 2 m
cos2 = 1
2
ml
m
or, sec2 = m
ml
2
1
or, tan2 +1 = m
ml
2
1
or, tan2 =m
mml
2
21
or, tan2 =m
l
2
1
tan = m
l
2
1 Ans.
78. (C) x sin a y cos a = 0
x sin a = y cos a
cos
sin=
x
y
xy3 + yx3 = xy
xy(y2 + x2) = xy
x2 + y2 = 1 Ans.
79. (C) 1 +
2= 90
2
= 90 1
In ABC, tan 1=
BD
AB
tan 1 =
x
AB... (i)
In ABC, tan 2 =
y
AB
tan(90 1) =
y
AB
cot 1 =
y
AB
tan 1 =
AB
y... (ii)
from equation (i) and (ii),
x
AB=
AB
y
(AB)2 = xy
AB = xy Ans.
80. (A)3x + 4y = 12
x = 0, y = 3 A(0, 3)
y = 0, x = 4 B(4, 0)
6y + 8y = 60
x = 0, y = 7.5 C(0, 7.5)
y = 0, x = 10 D(10, 0)
Area of Trapezium
= Area of COD Area of AOB
= 2
1 10 7.5
2
1 4 3
= 37.5 6 = 31.5 Ans.
81. (A) Curved surface area of a frustum
= ! (r + R)L
Here, R = 2
8= 4m, r =
2
4 = 2 m
and l = 22 )( rRh
l = 22 26 = 40
Curved surafce area
= 7
22 (2 + 4) 40
= 119.26 = 118.4 m2 (approx.) Ans.
82. (C) Required difference = (l b h) ! r2h
= 10 10 21 7
222
2
10
21
= 2100 1650 = 450 cm3 Ans.
83. (A)Total surface area of prism
= lateral surface area + 2 (area
of base)
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11
Here, s = 2
cba =
2
132021 = 27
Required area = (21 + 20 + 13) 30 + 2
)1327)(2027)(2127(27
= 54 30 + 2 147627
= 1620 + 2 126
= 1872 sq. m Ans.
84. (C) Curved surface area of cone = Area of
sector of circle
! rl = !R2 360
120
Here, l = R
r = 15 360
120 = 5 cm
h = 25225 = 210 cm
Volume of cone = 3
1! r2h
= 3
1 ! 25 210
= 2250 ! /3 cm3 Ans.
85. (B) The error percent in area
= + 5 6 100
65 = 1.3% Ans.
86. (D) Volume of the ball
= Volume of raised water
3
3
4r! = 2(12) (6.75)
r3 = 729r = 9 cm Ans.
87. (C) Let the diameter of third ball be 2r.
3
4!
3
2
3
=
3
4!
3 3 31.5 2 2
2 2 2
r
8
27 =
64
27 + 1 + r3
r3 = 8
27
64
27 1
r3 = 64
6427216 =
64
125
r = 4
5 cm
2r = 4
10= 2.5 cm
88. (D) Total number of smaller cubes
= 111
444
= 64
Increase in surface area
= 64 6 (1)2 6 (4)2
=384 96 = 288 cm2
Required percentage
= 96
288 100 = 300% Ans.
89. (B) Radius of the circle = 2
14 = 7 cm
Area of circle = 22
7 7 7 = 154 cm2 Ans.
90. (B) Volume of cube = 2
1 25 20 4
= 1000 cm3
Side of cube = 10 cm
Total surface area = 6 (10)2
= 600 cm2
91. (B) Ratio of processing cost of water for
industrial, energy and domestic usage
= 3 : 5 : 2
In 2006, water usage for industrial,
energy and domestic = 25, 26 and 35
trillion litres
In 2009, water usage for industrial,
energy and domestic
= 49, 35. 30 trillion litres.
Ratio of processing cost for abovementioned usage in 2006 to that in 2009
= 230535349
235526325
= 60175137
7013075
= 382
275= 0.72 Ans.
92. (A) Usage in energy related sector in 2006
= 26 trillion litres
Usage in energy related sector in 2009
= 35 trillion litres
Required percentage increase
= 35 26
26
100 = 34.6% Ans.
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12
93. (C) Required percentage
= 25303549490
490
100
= 629
490 100 = 77.90% Ans.
94. (D) Required difference
=
= 629
30 100
531
35 100
= 4.77% 6.59%
= 1.82% (Ignore ve) Ans.
95. (B) Percentage increase in usage from 2006
to 2009.
Domestic = 629
30 100
531
35 100
= 4.77 6.59 = 1.82%
[ignore ve]
Industrial = 629
49 100
531
25 100
= 7.79 4.71 = 3.08%
Others = 629
25 100
531
35 100
= 3.97 6.59 = 2.62%
[ignore ve]
Energy =629
25 100
531
26 100
= 5.56 4.89 = 0.66% Ans.
96. (C) One lakh Ans.
97. (A) Required factor = 400
320= 0.8 Ans.
98. (D) In 1990 it shows the least increasing
over the preceding year. Ans.
99. (A) Average sales
= 6
400440420400320340
= 6
2320 = 386.66
Sales are above average in 1988, 1989,
1990 and1991 and below average in 1986
and 1987
Required ratio = 4 : 2 = 2 : 1 Ans.
100. (A) Average sales from 1988 to 1991
= 4
400440420400
= 415 Ans.
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1. *
2. D
3. B
4. D
5. B
6. A
7. C
8. A
9. *
10. A
11. A
12. C
13. A
14. A
15. D
16. A
17. A
18. C
19. D
20. C
21. D
22. D
23. A
24. C
25. B
26. C
27. A
28. D
29. B
30. A
31. B
32. C
33. C
34. B
35. D
36. B
37. D
38. A
39. A
40. A
41. D
42. A
43. D
44. B
45. C
46. D
47. D
48. B
49. D
50. C
51. B
52. A
53. A
54. D
55. C
56. A
57. A
58. D
59. B
60. C
61. D
62. D
63. C
64. B
65. C
66. C
67. B
68. B
69. B
70. B
71. C
72. D
73. B
74. D
75. B
76. D
77. B
78. C
79. C
80. A
SSC MAINS (MATHS) MOCK TEST -13 (ANSWER SHEET)
81. A
82. C
83. A
84. C
85. B
86. D
87. C
88. D
89. B
90. B
91. B
92. A
93. C
94. D
95. B
96. C
97. A
98. D
99. A
100. A
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14
54. (C) In the given question please readpolynomial 2x4 3x3 2x2 + 6x + 2
as polynomial 2x4 3x3 3x2 + 6x 2 and according to the correct polynomial, the correct answer will be option (C).89. (*) The given solution is correct,
according to which the answer is 4000m3,which is present as both the options(A) & (B)
11. (C) Let the five consecutive odd nos. bex, x + 2, x + 4, x + 6 & x + 8.
So, ATQ,5x + 2 + 4 + 6 + 8 = 41 5
or, 5x = 205 20
x = 5
185= 37
So, product of 1st & last terms (A & E)= 37 (37 + 8) = 1665 Ans.
37. (B) In the solution given,Duration of time = 16 hours
(which is correct)and point of time = 3 pm + 16 hours
= 7 am Ans.42. (D) A + B + C = 15 3 = 45
& B + C + D = 16 3 = 48 D A = 3or, 19 A = 3 A = 16
CORRECTIONS FOR MOCK TEST - 12
So, The required % = %10016
16 = 100% Ans.
68. (D)or, x = 31
3
2
222
or, x 2 = 31
3
2
22
So,
" 3
1
3
2
3
1
3
23
3
12
3
2
3 222.2.322)2(x
or, (x 2)3 = 4 + 2 + 3.2(x2)or, (x 2)3 = 6 + 6(x 2)or, x3 6x2 + 12x 8 6 6x 12 = 0or, x3 6x2 + 6x 26 = 0
x3 6x2 + 6x = 26 Ans.
69.(*) (x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + zx)or, (1)2 = x2 + y2 + z2 + 2(1) x2 + y2 + z2 = 3 Ans.
76.(C) The given solution is totally correct. but the correct option is C (12 cm). Ans.
CORRECTIONS FOR MOCK TEST - 10
88.(A) Correct question should be "cos2 + 2cos " & according to this question the correct
option is A. Ans.
CORRECTIONS FOR MOCK TEST - 9
21.(B) Quantity of dry fruits (without water) in 100kg fresh fruit = 32% of 100 kg = 32 kg.So, required quantity of dry fruit (dry fruitcontains 80% fruit and 20% water)
=80
10032kg = 40 kg Ans.
46(B).
Now, C takes 1 hour 54 minutes = 114 minSo, time taken by B to cover the same
Distance = 114
3min = 38 min. Ans.
CORRECTIONS FOR MOCK TEST - 8
97. (D) Can not be determind as the populationis not known.
100.(D)Can not be determind as the populationis not known.